The arcade has a floor that is 120 feet by 75 feet. A scale drawing of the floor on grid paper uses a scale of 1 in: 5. Feet. Wh
1 answer:
Answer:
2 feet by 1.25 feet
Step-by-step explanation:
The arcade floor has dimensions 120 ft by 75 ft.
The scale of the drawing on grid paper is 1 in : 5 feet. This is a reduced scale. This means that the size of the arcade floor was reduced in the drawing.
Let us simply the ratio 1 in : 5 feet.
1 foot = 12 inches
=> 5 feet = 5 * 12 = 60 inches
The ratio can be rewritten as 1 in : 60 in = 1 : 60.
Hence, multiply the individual dimensions of the arcade floor by .
120 feet will be:
120 * = 2 feet
75 feet will be:
75 * = 1.25 feet
Therefore, the dimensions of the scale drawing are 2 feet by 1.25 feet.
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Let <em>a </em>(<em>n</em>) denote the <em>n</em>-th term of the given sequence.
Check the forward differences, and denote the <em>n</em>-th difference by <em>b </em>(<em>n</em>). That is,
<em>b </em>(<em>n</em>) = <em>a </em>(<em>n</em> + 1) - <em>a </em>(<em>n</em>)
These so-called first differences are
<em>b</em> (1) = <em>a</em> (2) - <em>a</em> (1) = 25 - 7 = 18
<em>b</em> (2) = <em>a</em> (3) - <em>a</em> (2) = 51 - 25 = 26
<em>b </em>(3) = <em>a</em> (4) - <em>a</em> (3) = 85 - 51 = 34
<em>b</em> (4) = <em>a </em>(5) - <em>a</em> (4) = 127 - 85 = 42
Now consider this sequence of differences,
18, 26, 34, 42, …
and notice that the difference between consecutive terms in this sequence <em>b</em> is 8:
26 - 18 = 8
34 - 26 = 8
42 - 34 = 8
and so on. This means <em>b</em> is an arithmetic sequence, and in particular follows the rule
<em>b</em> (<em>n</em>) = 18 + 8 (<em>n</em> - 1) = 8<em>n</em> + 10
for <em>n</em> ≥ 1.
So we have
<em>a </em>(<em>n</em> + 1) - <em>a </em>(<em>n</em>) = 8<em>n</em> + 10
or, replacing <em>n</em> + 1 with <em>n</em>,
<em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 8 (<em>n</em> - 1) + 10
<em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 8<em>n</em> + 2
We can solve for <em>a</em> (<em>n</em>) by iteratively substituting:
<em>a</em> (<em>n</em>) = [<em>a</em> (<em>n</em> - 2) + 8 (<em>n</em> - 1) + 2] + 8<em>n</em> + 2
<em>a</em> (<em>n</em>) = <em>a </em>(<em>n</em> - 2) + 8 (<em>n</em> + (<em>n</em> - 1)) + 2×2
<em>a</em> (<em>n</em>) = [<em>a</em> (<em>n</em> - 3) + 8 (<em>n</em> - 2) + 2] + 8 (<em>n</em> + (<em>n</em> - 1)) + 2×2
<em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 3) + 8 (<em>n</em> + (<em>n</em> - 1) + (<em>n</em> - 2)) + 3×2
and so on. The pattern should be clear; we end up with
<em>a</em> (<em>n</em>) = <em>a</em> (1) + 8 (<em>n</em> + (<em>n</em> - 1) + … + 3 + 2) + (<em>n</em> - 1)×2
The middle group is the sum,
so that
<em>a</em> (<em>n</em>) = <em>a</em> (1) + (4<em>n</em> ² + 4<em>n</em> - 8) + 2 (<em>n</em> - 1)
<em>a</em> (<em>n</em>) = 4<em>n</em> ² + 6<em>n</em> - 3